GCSE数学 联立方程 代入法 消元法 图像法
联立方程要求两个未知数同时满足两个条件，是 GCSE 数学代数部分少数在基础卷与高阶卷都会出现的考点。AQA 8300 与 Edexcel 1MA1 的官方 spec 都把解联立方程列在 Algebra 之下，高阶卷另外要求一道线性方程配一道二次方程的情形。真正要用的工具只有三件：代入、消元、画图，多数失分不是因为不会算，而是没有先判断该用哪一件，或者算完就交卷。三种方法在基础卷与高阶卷都会考，题号靠后的那几道通常要先把式子整理一遍才能动手算。下面把这三种方法各自拆开，例题按卷面要求写，过程分落在哪里也一并说明。
Simultaneous equations ask for two unknown values that satisfy two conditions at once, and the topic appears on both Foundation and Higher tier papers; the AQA 8300 and Edexcel 1MA1 specifications place it in the algebra section, and Higher tier adds a pair where one equation is linear and the other is quadratic. Only three tools are needed: substitution, elimination and graphs. Marks are lost for choosing the wrong tool or for stopping before the answer has been checked, not for the arithmetic itself. The questions near the end of a paper usually ask for a rearrangement before any calculation can start. The examples below follow the way papers phrase their questions, including where the method marks sit.
一、解的含义与三条判据
联立方程的一组解就是一对数值，写成 (x, y) 的形式，这一对必须放进两个方程都成立。两个线性方程的图像是两条直线，解对应两条线的交点坐标。斜率不同时只有唯一交点；斜率相同而截距不同时两条线平行，方程组没有解；斜率与截距都相同时两条线重合，解有无穷多组，答案一般写成 x 与 y 的关系式而不是具体数字。无解与无穷多解同样会出现在卷面上，判断时不必先解出具体数值，把两个方程都整理成 y = mx + c 再比较斜率与截距最快，斜率相同就去看截距，截距也相同才是同一条线。很多学生以为算出数字就算完成，其实第二个方程就在手边，代回去一步就能确认。OCR J560 的官方 spec 把图形解法与代数解法并列，所以读图题的过程分落在取点、连线与读出交点这三处。
A solution is one coordinate pair that makes both equations true at the same time; putting that pair back into each equation is the quickest check available. Two linear equations draw as two straight lines, so the solution is the crossing point: different gradients give one crossing, equal gradients with different intercepts give parallel lines with no solution, and identical lines give infinitely many solutions written as a relationship between x and y. The OCR J560 specification treats the graphical method as a technique in its own right, with marks for plotting the points and drawing the lines as well as for reading the intersection. Students who treat the second equation as decoration lose the easiest mark on the question.
二、代入法：从现成表达式入手
代入法适合其中一道方程已经把某个未知数单独写出来的情况，例如 y = 2x + 1，或者某个未知数的系数是 1。做法是把那个表达式整体塞进另一道方程，两元方程就降成一元方程，解出剩下的未知数，再回代求另一个。例题：解 y = 2x + 1 与 3x + y = 11。把 y 代入第二式得到 3x + (2x + 1) = 11，合并得 5x + 1 = 11，所以 x = 2，回代得 y = 5。把 (2, 5) 代回 3x + y 得到 6 + 5 = 11，与题给一致。代入之后如果出现分数，保留分数比转成小数更保险，最后统一约分；回代时要用原题里形式最简单的那道方程，避免把中间误差带进第二个未知数。常见的错误是忘了括号，方程 y = 3 - x 代入之后写成 3x + 3 - x，符号就翻了。题目不给现成表达式时，挑系数为 1 或 -1 的那个未知数改写，花一行换个好算的局面。
Substitution works when one equation already isolates an unknown, such as y = 2x + 1, or when one unknown has a coefficient of 1; the expression is placed into the other equation so that a single unknown is left, and its value is then substituted back. With y = 2x + 1 and 3x + y = 11 the substitution gives 3x + (2x + 1) = 11, so x = 2 and y = 5, and the pair checks in both equations. Dropped brackets cause most of the damage here, since writing 3x + 3 - x for 3x + (3 - x) flips a sign and the answer drifts; keep the brackets until the terms are collected. Fractions found partway through are safer left as fractions, and substituting back into the simpler equation keeps the second value accurate. If no expression is handed to you, rearranging the unknown with coefficient 1 or -1 first costs a line and saves several later.
三、消元法：把系数配平再加减
消元法用在两个方程里同一个未知数的系数成倍数关系的时候。系数相反就把两式相加，系数相同就把两式相减，一个未知数先消失，剩下的按一元方程处理。例题：解 2x + 3y = 17 与 3x + 2y = 18。第一式乘 3 得 6x + 9y = 51，第二式乘 2 得 6x + 4y = 36，两式相减得 5y = 15，所以 y = 3，回代 2x + 9 = 17 得 x = 4。检验 (4, 3)：3 × 4 + 2 × 3 = 18，与第二式一致。两个未知数的系数都配不平时，挑乘的次数更少的一边下手，把系数 2 与 4 都乘成 4 只要一次乘法，比都乘成 12 少一步。很多学生担心两边乘一个数会改变答案，只要每一项都乘同一个数，解集就不变；错得最多的是只乘了含未知数的项，把 2x + 3y = 17 写成 6x + 9y = 17，常数项被丢掉。相减时整式要加括号，符号问题与代入法同一回事。
Elimination suits a pair where the coefficients of one unknown are already linked; opposite signs mean adding the equations, matching signs mean subtracting them, and one unknown disappears before the single-variable equation is finished. Take 2x + 3y = 17 and 3x + 2y = 18: multiplying the first by 3 and the second by 2 gives 6x + 9y = 51 and 6x + 4y = 36, so subtracting leaves 5y = 15, hence y = 3 and x = 4, which checks as 3 × 4 + 2 × 3 = 18. Multiplying both sides by a number does not change the solution set, provided every term is multiplied; the error examiners see is a constant left behind, turning 2x + 3y = 17 into 6x + 9y = 17. Where neither pair of coefficients divides neatly, scale the smaller numbers, since turning 2 and 4 into 4 and 4 needs one multiplication fewer than a common multiple. Brackets matter again when the subtraction happens.
四、图像法：画两条线读交点
图像法要求把两个方程都整理成 y = mx + c 的形式，各取三到四个点列表描点，连成直线，交点的横纵坐标就是解。这部分在卷面上很少要求精确到小数，交点落在格点上时直接读出整数，落在两格之间就按题目要求的精度估读。斜率相同时两条线平行，图上没有交点，直接写无解，比硬算更省时间。表格里把每列的 x 与 y 对应写清楚，连点成线用直尺，手画的弧线会被当成取点错误；两条线交叉得很平缓时，交点坐标要按题目给的精度多估一位。很多学生以为图像法只是检查手段，题目写明 solve graphically 时却拿不出图，草算出的答案得不到分。取点时把 x 取成整数并让这些点跨过交点两侧，交点位置才不会跑到纸外。
The graphical method starts by rearranging both equations into y = mx + c, after which three or four points each give two straight lines and the crossing point is the solution. Precision is rarely demanded to a decimal; a crossing on a grid intersection can be read directly, while one between lines is estimated to the accuracy the question states. Parallel lines mean no solution and can be answered from the sketch, which saves the algebra entirely. A question that says solve graphically will not reward numbers that come with no graph, so the plotting is part of the answer rather than a check. Rulers keep the lines straight, and a shallow crossing needs one more decimal place than the table suggests. Choosing integer x values on both sides of the crossing keeps the intersection on the page.
五、应用题：先定义未知数再列式
应用题的第一步不是列方程，而是把两个未知数写清楚，例如设成人票 x 张、儿童票 y 张，并在答案里注明单位。列式的依据通常是两个独立条件，一个把数量相加，另一个给出总价或总量。例题：成人票 12 元、儿童票 8 元，共卖出 30 张票，收入 304 元。写出 x + y = 30 与 12x + 8y = 304，把第一式乘 8 得 8x + 8y = 240，两式相减得 4x = 64，所以 x = 16，y = 14。检验：16 × 12 + 14 × 8 = 192 + 112 = 304，与题干一致。列完两个方程先看一眼它们是否互相独立，如果第二式只是第一式两边乘同一个数得到的，条件实际上只有一个，答案不唯一，此时回头读题，通常还有一个条件没被用上。年龄型题目常见的问题是把往年的情形当作未知数的一部分，其中一个式子会用错年份；速度与距离型题目要统一单位，分钟与小时混用会让答案差 60 倍。答案必须回到题干问法，问人数就写人数，不要只留下 x 与 y。
Word problems are solved by naming the unknowns first, such as x for adult tickets and y for child tickets, and by keeping the units in the final answer. Two independent conditions then supply the equations, one usually adding quantities and the other adding money or totals. A concert selling 30 tickets at 12 yuan for adults and 8 yuan for children with 304 yuan taken gives x + y = 30 and 12x + 8y = 304, which elimination turns into x = 16 and y = 14, and the check is 16 × 12 + 14 × 8 = 304. Two equations that are multiples of each other carry only one piece of information, so one condition has been left unused and the answer will not be unique. Age questions go wrong when the earlier year is mixed into the wrong equation, and speed questions go wrong when minutes and hours are mixed without converting. Finish by answering the question asked, in words, rather than leaving x and y on the page.
六、线性与二次混合：代入是稳的那条路
高阶卷会出现一道线性方程配一道二次方程，消元法在这里帮不上忙，代入是稳定的做法。例题：解 y = x + 1 与 x² + y² = 25。把 y 代入第二式得到 x² + (x + 1)² = 25，展开为 2x² + 2x + 1 = 25，整理成 x² + x - 12 = 0，因式分解 (x + 4)(x - 3) = 0，所以 x = 3 或 x = -4，对应的 y 分别是 4 与 -3。整理出二次方程之后，题目问的是交点坐标，两对数字都要写全，少写一组按答案点数通常扣掉一半分。两个交点 (3, 4) 与 (-4, -3) 都要写出来，只写一组是这一步最常见的失分点。整理后的二次方程若无实根，说明直线与圆或直线与曲线没有交点，答案写无解。题目写 hence solve 时，前一问的结果要用上，另起一套算法反而会被扣分。
Higher tier papers pair one linear equation with one quadratic, and elimination cannot do the job there, so substitution is the dependable route. For y = x + 1 and x² + y² = 25, substituting gives x² + (x + 1)² = 25, which becomes x² + x - 12 = 0 and factorises as (x + 4)(x - 3) = 0; the pairs are (3, 4) and (-4, -3), and both belong in the answer. A quadratic with no real roots means the line and the curve never meet, so the result is no solution. When a question says hence, the earlier part has to be used, and starting again from scratch often costs the marks that were available.
七、指令词与卷面写法
卷面上的指令词决定了答案要写多深。solve 要求给出 x 与 y 的值，show that 要求把推导过程写全，hence 要求接着上一问的结论往下走。一分的简单题只给答案也不算错，多分题的过程分单独给，写消元时把乘过之后的两个方程都抄一遍，回代那一步写清楚用的是哪条方程。用不上的草稿不要抄进卷面，直接写最终方程组更容易被读懂。分数答案不必转成小数，3/2 这种形式在 GCSE 里完全接受。把每一步写在左侧、答案写在右下角，复查时一行一行往下看，比在草稿纸上翻找快得多。答案写完后核对一遍，代入另一个方程比重新算一遍快。阅卷时见到最多的情况不是算错，而是只写了 x 的值就停笔，y 被留在草稿纸上。
Command words set the depth of the answer: solve wants the values of x and y, show that wants the full derivation, and hence wants the earlier result carried forward. Short one-mark questions accept a bare answer, while longer ones carry separate method marks, so write out both equations after scaling and state which equation the value was put back into. Answers as fractions are accepted, so a value such as 3/2 needs no conversion to a decimal. Laying the working down the left of the page and the answers at the bottom right makes the check quicker to read. Checking by substituting into the other equation is faster than repeating the arithmetic. Candidates who stop after finding x and leave y in the margin lose marks that were already earned.
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